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Stats: 1061763 members, 1233069 topics. Date: Tuesday, 21 May 2013 at 07:19 PM
|The Door-Goat-Car Logic Problem by delomos: 4:01am On Mar 30, 2012|
I ran into this probability issue yesterday and it's quite interesting, I'd like to read some take on it:
There are three doors in front of you, one has a car, and the other two have goats behind them. You been promised you will win the car behind one of the doors if you can can guess the correct door with the car behind it.
The person opening the door to reveal your choice knows which of the doors has the car. You made your guess.
The person opens the door, there is a goat behind it, he then asked: "will you want to guess again or stick to your original choice before I open the next door?" What will be your response and what's the rationale?
Now there are five hundred doors all with goats expect one with a car, every thing remains the same (mostly that the person opening the door know which has the car behind is), after opening the 498th door, he asked if you will like to change your mind mind or stick with your original choice. What's your choice/rationale?
*extra points for those who know what area of computer science this conundrum applies
|Re: The Door-Goat-Car Logic Problem by okeyxyz(m): 10:16am On Mar 30, 2012|
he then asked: "will you want to guess again or stick to your original choice before I open the next door?"
the above suggests that the previous choice\door which turns out to be wrong is put back into the pool of choices\doors, the contents are shuffled, before allowing you to make another choice, so changing your mind\door or sticking with previous door does nothing to improve your chances of guessing the correct content(car). so, there's no logical way to better the odds here!
|Re: The Door-Goat-Car Logic Problem by delomos: 11:36am On Mar 30, 2012|
@okeyxyz: see the pattern in Story 1 & Story 2, ' that the previous choice\door which turns out to be wrong is put back into the pool of choices\doors", why isn't it taken out?
|Re: The Door-Goat-Car Logic Problem by codeaddict(m): 10:04pm On Mar 30, 2012|
Before choosing a door, his probability of choosing the door with a car was 1/3, by choosing to switch doors, he increases the likelihood of choosing the door with a car to 2/3. Mathematically, it might be advantageous to choose another door.
|Re: The Door-Goat-Car Logic Problem by delomos: 4:07am On Mar 31, 2012|
@codeaddict: Nice thinking... Does the same apply to Story 2?
|Re: The Door-Goat-Car Logic Problem by codeaddict(m): 1:56pm On Mar 31, 2012|
Yeah, i believe so-mathematically.
|Re: The Door-Goat-Car Logic Problem by lordZOUGA(m): 7:04pm On Mar 31, 2012|
in both cases... he should close the door and make another choice... because, if the person opening the door knows the position of the car then it means the position of the car is a constant and is not modified at runtime. closing the door grants him a higher probability (assuming he progresses linearly and doesn't open the same door twice). errrrrrm... I'd say that this (to me) reminds me more of the uncertainty principle in quantum mechanics where it is said that objects can exist simultaneously in a given number of states(constraints) until forced by an external body to choose a stable state(saying it loosely). so before any door is opened, what is behind the door is a car and a goat. jus saying
|Re: The Door-Goat-Car Logic Problem by lordZOUGA(m): 7:04pm On Mar 31, 2012|
was my analogy correct??
|Re: The Door-Goat-Car Logic Problem by delomos: 8:49pm On Mar 31, 2012|
lordZOUGA: in both cases... he should close the door and make another choice...That in facts creates a new problem of an infinite loop, again, don't forget the fact that the person opening the door knows which door has the car -- re-work the rationale
lordZOUGA:You're thinking of Schrödinger's cat, this isn't exactly the case
lordZOUGA:This is only true for the person wanting to win the car, again, the other person knows what's precisely behind every door.
codeaddict: Yeah, i believe so-mathematically.How would you rationale this mathematically. Given that (in Story 1) , Pr(guess 1) = 1/3 ?
|Re: The Door-Goat-Car Logic Problem by jetsetgo: 9:27pm On Mar 31, 2012|
Well, in the first instance the probability was 1 in 3 or 33.33%, opening a wrong door did not increase the possibility to 50% it went further because by changing the original choice selected you have moved from 1 choice out of 3 to 2 out of the original 3 and have thus derived 2/3 probability or a percentage of 66.67%.
Applying same logic to the case 2 produces linear relation that is slightly decelerated before stabilising.
|Re: The Door-Goat-Car Logic Problem by Fayimora(m): 10:36pm On Mar 31, 2012|
|Re: The Door-Goat-Car Logic Problem by Fayimora(m): 10:38pm On Mar 31, 2012|
When you see questions like this, I have learn't that there is an 85% probability that the obvious way isin't the right answer.
|Re: The Door-Goat-Car Logic Problem by lordZOUGA(m): 11:22pm On Mar 31, 2012|
@delomos,I meant that if the contents behind the door is not shuffled(not modified at runtime) then anytime he recants his choice, the probability that his next choice will hold the car increases... but if it is being shuffled, it doesn't matter if he recants or not, the probability remains the same. if the game moderator knows the position of the car then its not being modified -> a closed system.
|Re: The Door-Goat-Car Logic Problem by delomos: 1:07am On Apr 01, 2012|
Fayimora:Slowly drawing to a truism, expatiate on how Bayes would attack this... (or come to a conclusion with higher than average error bar)
I'm assuming "recants" means "change"?
Remove the "shuffling" -- per the story, there is no chance of shuffling (the person opening the door only knows what's behind the door not having any action on it).
|Re: The Door-Goat-Car Logic Problem by Fayimora(m): 1:16am On Apr 01, 2012|
delomos: @FayimoraLol its really not that obvious!!!
|Re: The Door-Goat-Car Logic Problem by delomos: 3:30am On Apr 01, 2012|
Fayimora:So get back to the drawing board and let's see how you'd re-think it.... or expand on this...
|Re: The Door-Goat-Car Logic Problem by okeyxyz(m): 7:12am On Apr 01, 2012|
if there is no reshuffling, then the problem scenario is faulty. why else would the "game moderator" ask:
"will you want to guess again or stick to your original choice before I open the next door?"why would the player want to stick with his original choice if it is no longer in the equation? And, the only way to effectively remain in the equation is that it's effect can not be discounted thus necessitating a reshuffle,
|Re: The Door-Goat-Car Logic Problem by codeaddict(m): 11:49am On Apr 01, 2012|
delomos:working from this story, switching doors will be unfavourable only if he chose a car at first. The probability of this occurring is 1/500, therefore switching will be more favourable because the probability is 499/500 [1-(1/500)].
Does the person opening the door open the first 498 doors or the 498th door?
|Re: The Door-Goat-Car Logic Problem by okeyxyz(m): 12:29pm On Apr 01, 2012|
No, the probability would be 1/499, that is assuming there's no returning the previous choice into the poll & reshuffling after the first choice, otherwise it remains 1/500.
suppose there's no return & reshuffle, the probabilities in story_1 and story_2 are the same: 1/2.
|Re: The Door-Goat-Car Logic Problem by jetsetgo: 12:02pm On Apr 03, 2012|
Whats the basis of this blah blah blah and more blah blah?
@ First, me thought it was gon' lead somewhere see!
me was thinking we's gon' learn something from this see!
But it seems you guys are just bored and want ta flex ya opolo prowess to one another.
interesting at first but kinda deadweight @ the moment so please just move on to the point and if there is none.... Get a girlfriend or a hobby guys
|Re: The Door-Goat-Car Logic Problem by delomos: 2:54pm On Apr 03, 2012|
the first 498th door through to 498th -- very excellent catching the difference, you've already nailed it but more will be appreciated.
@okeyxyz: read the previous analysis of @codeaddict, even if you don't agree with the conclusion (which in my opinion has a higher confidence level than yours), there's somethign to be learned from the approach.
@jetsetgo: usually I will ignore this but people like you are why Nigeria is lacking innovation, if you're too lazy to think, go like peoples picture on facebook, or follow celebrities on twitters; if that doesn't do you, go and troll lackadaisicals like you on notjustok.com.
When you can understand a single post here: http://cstheory.stackexchange.com/ come back and let's talk.
I like you advise, yes I do have a girlfriend who you can't match a quarter of her wit and I'm also a music hobbyist (I play about 5 instruments bro) -- computing feeds me and it's my life. Now you -- go get a life.
|Re: The Door-Goat-Car Logic Problem by okeyxyz(m): 4:22pm On Apr 03, 2012|
|Re: The Door-Goat-Car Logic Problem by okeyxyz(m): 4:28pm On Apr 03, 2012|
OK, i see where you are coming from, the clarification has finally been made. which takes us back to my earlier objection that the question itself is faulty because of ambiguity, and as a programmer\computer scientist you know that computing(unlike mathematics) does not tolerate ambiguity, it's a sure recipe for errors. good try though
|Re: The Door-Goat-Car Logic Problem by delomos: 6:51pm On Apr 03, 2012|
I think you have it the other way around, scientist are faced with ambiguities, then they try to "make it fit", hence does who can properly deal with ambiguity get a hang of things better. In fact, abstracts drive the industry (trying to continually model things to the uncertainty of the real whole); after all the field of [Computing] Combinatorial is largely based on understanding ambiguity or making them -- but I digress.
How can the question be faulty? Think of it as a problem being solved, you never have all the "facts" or "variables" when you're presented with a problem -- it's you're job as the solver to see the gaps and place them. Again think of Occam's Razor and put this question in that context -- [?]
|Re: The Door-Goat-Car Logic Problem by okeyxyz(m): 7:42pm On Apr 03, 2012|
you misunderstand my point, but taking from your words: then they try to "make it fit" implies they try to eliminate these ambiguities right? in other words they don't tolerate ambiguities, rather they look for definitions, precisions, validities, scopes etc.
i don't need all the facts and variables, but i do need your POV to understand what i'm dealing with. this is what codeaddict has managed to elicit from you, thus half of the problem is solved already.
|Re: The Door-Goat-Car Logic Problem by ekt_bear: 8:17pm On Apr 03, 2012|
The Monty Hall problem. Yeah I found this quite counterintuitive the first time I saw it. I think it helps a lot to think of 1 billion doors rather than just 3.
|Re: The Door-Goat-Car Logic Problem by delomos: 12:56am On Apr 04, 2012|
My apologies, I did misunderstand your point. As @ekt_bear stated, yes, it's the Monty Hall problem and it's terribly easy to get the problem wrong (Almost like dating a chic who "says" she's a virgin) -- and again, as @ekt_bear stated, I agree that it's easier to solve the problem when one thinks of 1 billion doors instead of three (actually that's why I added story 2, the problem has just one day)
In light of @ekt_bear's exposition, @ekt_bear and @okeyxyz how do you approach it?
How do you approach it if this where a computational problem?
P.S It will be exciting! to read people thought on how this problem relates to a girl "saying" she's a virgin
|Re: The Door-Goat-Car Logic Problem by queensmith: 1:02am On Apr 04, 2012|
in both stories don't you end up with a 50% chance of getting the correct door? There are only 2 doors left? one with a car amd the other with a goat? correct me if i'm wrong.
you might as well stick with the original guess because there is an equal chance of it being wrong as it being right.
|Re: The Door-Goat-Car Logic Problem by ekt_bear: 1:21am On Apr 04, 2012|
So imagine 1 billion doors. All but one of them has a goat in it. One of them has a car.
Let us assume that the door containing the car is chosen uniformly at random, and you have no knowledge of which door.
So you pick a door. You only have a 1 in 1 billion chance of being correct, right? So you are almost certainly wrong.
Now, the host eliminates 9,999,998 doors with goats in them. So you can keep your door, or switch to the other one he has. So the question is, what is the probability of you winning if you switch?
Or equivalently, what is the probability of me LOSING if I switch? I only lose if I switch if I'd picked the right door to start. But this only happens with probability 1 in 1 billion. So this means the probability of me WINNING if I switch is pretty close to 1.
We can run through this same exercise with 3 doors and get:
Prob(Lose if switch) = 1/3
Prob(Win if switch) = 2/3
Prob(Win if don't switch) = 1/3
Prob(Lose if don't switch) = 2/3
Then I maximize my probability of winning by switching (2/3 is bigger than 1/3)
|Re: The Door-Goat-Car Logic Problem by queensmith: 1:27am On Apr 04, 2012|
how do you end up with 2/3? if you switch to one of the doors left won't it be left to 1/2? i dont get it *cries*
|Re: The Door-Goat-Car Logic Problem by ekt_bear: 1:31am On Apr 04, 2012|
Which 2/3s? There are two. Which one are you asking about?
|Re: The Door-Goat-Car Logic Problem by ekt_bear: 1:33am On Apr 04, 2012|
I win if I switch only if I picked the wrong door to begin with. I pick the wrong door with prob 2/3, right?
I win if I don't switch only if I picked the right door to begin with. This only happens with prob 1/3.
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